# Question #be972

Oct 17, 2016

Let the distance between p and q be S m and the accelaration of the particle is $\text{ "a} \frac{m}{s} ^ 2$. Let the total time to cover distance S is T s.

The particle starts with initial velocity $u = 0$ and describes 2 m in 1st sec.

So applying kinematic equation
$s = u t + \frac{1}{2} a {t}^{2}$
$2 = 0 \times 1 + \frac{1}{2} \times a \times {1}^{2}$
$\implies a = 4 \frac{m}{s} ^ 2$

Again $S = 0 \times T + \frac{1}{2} \times 4 \times {T}^{2}$

$\implies S = 2 {T}^{2}$

Distance covered in $\left(T - 3\right)$s

${S}_{\left(T - 3\right)} = 0 \times \left(T - 3\right) + \frac{1}{2} \times 4 {\left(T - 3\right)}^{2} = 2 {\left(T - 3\right)}^{2}$

Si the distance covered in last 3 sec

$S - {S}_{\left(T - 3\right)} = \frac{11}{16} S = 2 {T}^{2} - 2 {\left(T - 3\right)}^{2}$

$\implies \frac{11}{16} \times 2 {T}^{2} = 2 {T}^{2} - 2 {\left(T - 3\right)}^{2}$

$\implies {\left(T - 3\right)}^{2} = \frac{5}{16} {T}^{2}$

$\implies \left(T - 3\right) = \frac{\sqrt{5}}{4} T$

$\implies \frac{4 - \sqrt{5}}{4} T = 3$

$\implies T = \frac{12}{4 - \sqrt{5}} s = \frac{12}{11} \left(4 + \sqrt{5}\right) s$

So the dustance between p and q

$S = 2 {T}^{2} = {12}^{2} / {11}^{2} {\left(4 + \sqrt{5}\right)}^{2} m = 46.28 m$